Answer:
Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O
The reactants are present in equimolar amounts, so there is no excess or limiting reactants.
(0.0500 L) x (0.600 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0300 mol H2O
(4.184 J/g·°C) x (50.0 g + 50.0 g) x (25.82 - 21.73)°C = 1711.256 J
(1711.256 J) / (0.0300 mol H2O) = 57042 J/mol = 57.0 kJ/mol H2O
Explanation: